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Balance Cl2 plus KBr - Br2 plus KCl?
The balanced equation for Cl2 + 2KBr -> 2Br2 + 2KCl is balanced as it conserves the number of atoms on both sides of the reaction. Two moles of KBr reacts with one mole of Cl2 to produce two moles each of Br2 and KCl.
What is the answer of this cl2 plus kbr-kcl plus brz?
The correct chemical equation for the reaction is: Cl2 + 2KBr → 2KCl + Br2. The reaction involves chlorine gas (Cl2) reacting with potassium bromide (KBr) to form potassium chloride (KCl) and bromine gas (Br2).
For the reaction Cl2 plus 2KBr equals Br2 calculate the percentage yield if 200 g of Cl2 react with excess KBr to produce 410 g of Br2?
To calculate percentage yield, first determine the theoretical yield using the stoichiometry of the reaction: 1 mol Cl2 produces 1 mol Br2, so 200 g Cl2 is equivalent to 3.125 mol Br2. The molar mass of Br2 is 159.808 g/mol, so the theoretical yield is 159.808 g/mol * 3.125 mol = 499.4 g. Then, divide the actual yield (410 g) by the theoretical yield (499.4 g) and multiply by 100 to get the percentage yield: (410 g / 499.4 g) * 100 = 82%.
Write balanced equations for KCl and Br2?
KCl: 2K(s) + Cl2(g) -> 2KCl(s) Br2: Br2(l) -> 2Br(s)
Chemical equation for aluminum bromide plus chlorine yield aluminum chloride plus bromine?
Aluminum bromide (AlBr3) + Chlorine (Cl2) → Aluminum chloride (AlCl3) + Bromine (Br2)
Balance Cl2 plus KBr - Br2 plus KCl?
The balanced equation for Cl2 + 2KBr -> 2Br2 + 2KCl is balanced as it conserves the number of atoms on both sides of the reaction. Two moles of KBr reacts with one mole of Cl2 to produce two moles each of Br2 and KCl.
What is the answer of this cl2 plus kbr-kcl plus brz?
The correct chemical equation for the reaction is: Cl2 + 2KBr → 2KCl + Br2. The reaction involves chlorine gas (Cl2) reacting with potassium bromide (KBr) to form potassium chloride (KCl) and bromine gas (Br2).
For the reaction Cl2 plus 2KBr equals Br2 calculate the percentage yield if 200 g of Cl2 react with excess KBr to produce 410 g of Br2?
To calculate percentage yield, first determine the theoretical yield using the stoichiometry of the reaction: 1 mol Cl2 produces 1 mol Br2, so 200 g Cl2 is equivalent to 3.125 mol Br2. The molar mass of Br2 is 159.808 g/mol, so the theoretical yield is 159.808 g/mol * 3.125 mol = 499.4 g. Then, divide the actual yield (410 g) by the theoretical yield (499.4 g) and multiply by 100 to get the percentage yield: (410 g / 499.4 g) * 100 = 82%.
What is the predicted yield for Cl2 KBr--KCl Br2?
cl2 kbr---kcl br2i think u mean balance it right ^^;here u have cl2 kbr---kcl br2 so what u do iscl2 kbr---kcl br2cl=2 cl=1k=1 k=1br=1 br=2so u need to balance that ...u need to add (2) to kbr and add (2) to kcl so that u have Cl2 (2)KBr ----(2)KCl Br2 hope that will help ^^ so now u have them balanced by adding 2 in front of kbr that means u r multiplying them by 2 so that K is going to be k=2 and br is going to be br=2 and u r doing that because there is br =1 on one side and the other side there is br=2 and u need them balanced. By adding 2 to kcl means that the other k now is k=2 so as cl. hope its helpful ^____~
Write balanced equations for KCl and Br2?
KCl: 2K(s) + Cl2(g) -> 2KCl(s) Br2: Br2(l) -> 2Br(s)
What does BaCl2 plus F2 yield?
When barium chloride (BaCl2) reacts with fluorine (F2), it forms barium fluoride (BaF2) and chlorine gas (Cl2). The balanced chemical equation for this reaction is: BaCl2 + F2 -> BaF2 + Cl2.
How do you balance this equation K plus Br2--- KBR?
2 K + Br2 -> 2 Kbr
Is Cl2 plus Cl2 London bond?
Cl2 is non polar.So there are london bonds.
Write the equation for K plus Cl2?
2K + Cl2 --> 2KCl
In the reaction Mg plus Cl2 MgCl2 what should be the coefficient for Cl2?
1
Chemical equation for aluminum bromide plus chlorine yield aluminum chloride plus bromine?
Aluminum bromide (AlBr3) + Chlorine (Cl2) → Aluminum chloride (AlCl3) + Bromine (Br2)
What is the percent yield for Na plus Cl2 -- 2NaCl?
Percentage yield is given as:-%yield = (mass of product obtained experimentally/theoretical yield)x100where the theoretical yield is the maximum mass which could possibly be obtained assuming 100% stoichiometry.So, say as an example if you start with 71g of Cl2 then:-mol = mass/Mr Mr of Cl2 = 71mol = 71/71mol = 1 of Cl2This is the limiting reagent, assuming we have at least 46g of sodium, thus we double this number as the reaction is 1:21 mole Cl2---->2moles NaClagain we use:-mole = mass/Mr Mr of NaCl = 58.52 x Mr = mass2 x 58.5 = massmaximum mass of NaCl possible = 117gIf the reaction is carried out and we achieve 100g of NaCl then:-%yield = (100/117) x 100= 85.5%
